SOLUTION: The perimeter of a rectangular lot is at most 28m. The difference between the length and the width is at least 4m. What could be the possible dimensions of the lot?

Algebra ->  Rectangles -> SOLUTION: The perimeter of a rectangular lot is at most 28m. The difference between the length and the width is at least 4m. What could be the possible dimensions of the lot?      Log On


   

Question 826832: The perimeter of a rectangular lot is at most 28m. The difference between the length and the width is at least 4m. What could be the possible dimensions of the lot?
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
x length
w width
Perimeter, 2x%2B2w=28, simplifies to x%2Bw=14.

x-w%3E=4.

Solve for either variable from the equation and substitute it into the inequality.
w=14-x;
x-%2814-x%29%3E=4
x-14%2Bx%3E=4
2x-14%3E=4
x-7%3E=2
highlight%28x%3E=9%29
-
Note that x%3C14 because of perimeter restriction. This means, highlight%289%3C=x%3C14%29. You can determine the corresponding sizes for w.
w=14-x, so 14-9=5. This would mean, highlight%285%3E=w%3E0%29.
Those intervals for x and w allow an infinite set of pairs.
One example is 9 by 5. Another example is 13 by 1.